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Image for Exercise 10.20 A string is wrapped several times around the rim of a small hoop with radius 8.00 cm and mass

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Answer #1

Suppose after descending 65 cm , speed of centre is v.

then angular speed w = v/r

at that point total KE = translational KE + rotational KE

       = m v^2 /2 + Iw^2 /2   = (m v^2 /2 ) + ( (m r^2)(v/r)^2 /2)

KE = mv^2 /2 + mv^2 /2   = mv^2


using energy conservation,

PEi + KEi = PEf + KEf


mgh + 0 = 0 + mv^2


9.81 x 0.65 = v^2

v = 2.52 m/s ...........Ans ( B)


A) w = v/r = 2.52 / 0.08 = 31.56 rad/s

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